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  Doubts about Chern-Simons state as a solution of the Hamiltonian constraint in quantum gravity

+ 2 like - 0 dislike

I've been doing some work with both Baez's *Knots, gauge fields and gravity* (1) and Gambini, Pullin's *Loops, knots, gauge Theories and quantum gravity* (2), lately.

I have basically two problems: I understand that, in the ADM formalism, the Lagrangian density for the cosmological term of Einstein equation is given by
$$ L= q\Lambda \underline{N},$$
where $q$ is the determinant of the 3-metric, $\Lambda$ is the cosmological constant, and $\underline{N}$ is $q^\frac{-1}{2}$N (the lapse function). Also, that
are the densitized triads of the Ashtekar formalism. However, I don't get why $q$ can be given by the expression (7.53) from (2):
$$q=\frac{1}{6}\underline{\epsilon_{abc}}\epsilon^{ijk}\tilde{E^a_i}\tilde{E^b_j}\tilde{E^c_K}. $$
Is there a way to obtain such expression?

The second problem is: after promoting the Ashtekar variables to operators ( $\hat{A^a_i}$ and $\hat{E^a_i}=\frac{\delta}{\delta A^a_i}$ ), it's can be shown, for the Chern-Simons state
$$\psi_{\Lambda}= e^{-\frac{6}{\Lambda}S_{CS}},$$
with $S_{CS}$ being the Chern-Simons action
$$S_{CS}= \int_{\Sigma} tr \,(A\wedge dA +\frac{2}{3}A\wedge A\wedge A),$$
\frac{\delta}{\delta A^i_a}\psi_\Lambda = \frac{3}{\Lambda}\overline{\epsilon^{abc}}F^i_{bc}\psi_\Lambda \\
\underline{\epsilon_{abc}}\frac{\delta}{\delta A^i_a}\psi_\Lambda = \frac{6}{\Lambda}F^i_{bc}\psi_\Lambda,
which comes from expressions (7.70) and (7.71) from (2).

My problem is with the second line. Am I supposed to take 

$$ \underline{\epsilon_{abc}}\overline{\epsilon^{abc}} = 2~?$$  Why would that be true?

Sorry for the lengthy post. I'd be glad if someone could help me with these. 

asked Apr 27, 2016 in Theoretical Physics by Theoretician (10 points) [ no revision ]

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